a kiwi is launched at 15 m/s at an angle of 40 above horizontal, the question is in details?

Question

a) what is its speed when it is at maximum height?
b) how far the launch site will kiwi strike the ground?
c) what will be its speed upon impact upon impact with the ground?

Answer 1

a)the speed at apogee is all horizontal, which is
vx(t)=15*cos(40)

11.5 m/s

b) use
vy(t)=15*sin(40)-g*t

when vy(t)=0, the kiwi is at apogee and 1/2 the way through total flight time
t=15*sin(40)/9.8

at impact t=30*sin(40)/9.8

plug that into x(t)
where
x(t)=15*cos(40)*t
x(t)=15^2*sin(80)/9.8
14.76 m

c) The impact speed is the same as takeoff 15 m/s

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